Home > Documents > 17 Super Elevation and Spiral Curves

17 Super Elevation and Spiral Curves

Date post: 14-Oct-2014
Category:
Author: marwannaseem
View: 269 times
Download: 2 times
Share this document with a friend
Embed Size (px)
Popular Tags:
of 52 /52
1 Superelevation and Spiral Superelevation and Spiral Curves Curves CE 453 Lecture 18 CE 453 Lecture 18
Transcript

Superelevation and Spiral CurvesCE 453 Lecture 18

1

Objectives1. Define superelevation runoff length and methods of attainment (for simple and spiral curves) 2. Calculate spiral curve length

2

Other Issues Relating to Horizontal Curves1. 2. Need to coordinate with vertical and topography Not always neededMAXIMUM CENTERLINE DEFLECTION NOT REQUIRING HORIZONTAL CURVE Design Speed, mph 25 30 35 40 45 50 55 60 65 70 Maximum Deflection 530' 345' 245' 215' 115' 115' 100' 100' 045' 045'

Source: Ohio DOT Design Manual, Figure 202-1E

3

Attainment of Superelevation General1. 2. Tangent to superelevation Must be done gradually over a distance without appreciable reduction in speed or safety and with comfort Change in pavement slope should be consistent over a distance Methods (Exhibit 3-37 p. 186)a. b. c. Rotate pavement about centerline Rotate about inner edge of pavement Rotate about outside edge of pavement

3. 4.

4

Superelevation Transition Section Tangent Runout Section + Superelevation Runoff Section

5

Tangent Runout Section Length of roadway needed to accomplish a change in outside-lane cross slope from normal cross slope rate to zero

For rotation about centerline

6

Superelevation Runoff Section Length of roadway needed to accomplish a change in outside-lane cross slope from 0 to full superelevation or vice versa For undivided highways with crosssection rotated about centerline

7

Source: A Policy on Geometric Design of Highways and Streets (The Green Book). Washington, DC. American Association of State Highway and Transportation Officials, 2001 4th Ed.

8

Source: A Policy on Geometric Design of Highways and Streets (The Green Book). Washington, DC. American Association of State Highway and Transportation Officials, 2001 4th Ed.

9

10

Source: CalTrans Design Manual online, http://www.dot.ca.gov/hq/oppd/hdm/pdf/chp0200.pdf

11

Same as point E of GB

12Source: Iowa DOT Standard Road Plans

Superelevation must be attained over a length that includes the tangent and the curve 2. Typical: 66% on tangent and 33% on curve of length of runoff if no spiral 3. Iowa uses 70% and 30% if no spiral 4. Super runoff is all attained in Spiral if used (see lab manual (Iowa Spiral length = Runoff length)13

1.

Attainment Location WHERE

Minimum Length of Runoff for curve Lr based on drainage and aesthetics rate of transition of edge line from NC to full superelevation traditionally taken at 0.5% ( 1 foot rise per 200 feet along the road) current recommendation varies from 0.35% at 80 mph to 0.80% for 15mph (with further adjustments for number of lanes) 14

Minimum Length of Tangent RunoutLt = eNC x Lr ed

where eNC = normal cross slope rate (%) ed = design superelevation rate Lr = minimum length of superelevation runoff (ft) (Result is the edge slope is same as for Runoff segment)15

Length of Superelevation Runoffr

= multilane adjustment factor Adjusts for total width16

Relative Gradient (G) Maximum longitudinal slope Depends on design speed, higher speed = gentler slope. For example: For 15 mph, G = 0.78% For 80 mph, G = 0.35% See table, next page

17

Maximum Relative Gradient (G)

Source: A Policy on Geometric Design of Highways and Streets (The Green Book). Washington, DC. American Association of State Highway and Transportation Officials, 2001 4th Ed.

18

Multilane Adjustment

Runout and runoff must be adjusted for multilane rotation. See Iowa DOT manual section 2A-2 and Standard Road Plan RP-2

19

Length of Superelevation Runoff ExampleFor a 4-lane divided highway with crosssection rotated about centerline, design superelevation rate = 4%. Design speed is 50 mph. What is the minimum length of superelevation runoff (ft) Lr = 12e G 20

Lr = 12e = (12) (0.04) (1.5) G 0.005 Lr = 144 feet

21

Tangent runout lengthExample continued Lt = (eNC / ed ) x Lr

as defined previously, if NC = 2% Tangent runout for the example is: LT = 2% / 4% * 144 = 72 feet

22

From previous example, speed = 50 mph, e = 4% From chart runoff = 144 feet, same as from calculation

Source: A Policy on Geometric Design of Highways and Streets (The Green Book). Washington, DC. American 23 Association of State Highway and Transportation Officials, 2001 4th Ed.

Spiral Curve Transitions

24

Spiral Curve Transitions Vehicles follow a transition path as they enter or leave a horizontal curve Combination of high speed and sharp curvature can result in lateral shifts in position and encroachment on adjoining lanes25

Spirals1. Advantages a. Provides natural, easy to follow, path for drivers (less encroachment, promotes more uniform speeds), lateral force increases and decreases gradually b. Provides location for superelevation runoff (not part on tangent/curve) c. Provides transition in width when horizontal curve is widened d. Aesthetic26

Minimum Length of SpiralPossible Equations: Larger of (1) L = 3.15 V3 RC Where: L = minimum length of spiral (ft) V = speed (mph) R = curve radius (ft) C = rate of increase in centripetal acceleration (ft/s3) use 1-3 ft/s3 for highway)

27

Minimum Length of SpiralOr (2) L = (24pminR)1/2

Where: L = minimum length of spiral (ft) R = curve radius (ft) pmin = minimum lateral offset between the tangent and circular curve (0.66 feet)

28

Maximum Length of Spiral Safety problems may occur when spiral curves are too long drivers underestimate sharpness of approaching curve (driver expectancy)

29

Maximum Length of SpiralL = (24pmaxR)1/2 Where: L = maximum length of spiral (ft) R = curve radius (ft) pmax = maximum lateral offset between the tangent and circular curve (3.3 feet)

30

Length of Spiralo AASHTO also provides recommended spiral lengths based on driver behavior rather than a specific equation. See Table 16.12 of text and the associated tangent runout lengths in Table 16.13. o Superelevation runoff length is set equal to the spiral curve length when spirals are used. o Design Note: For construction purposes, round your designs to a reasonable values; e.g. Ls = 147 feet, round it to Ls = 150 feet.

31

Source: Iowa32 DOT Design Manual

33 Source: Iowa DOT Design Manual

Source: Iowa 34 DOT Design Manual

SPIRAL TERMINOLOGY

Source: Iowa DOT Design Manual

35

Attainment of superelevation on spiral curvesSee sketches that follow: Normal Crown (DOT pt A) 1. Tangent Runout (sometimes known as crown runoff): removal of adverse crown (DOT A to B) B = TS 2. Point of reversal of crown (DOT C) note A to B = B to C 3. Length of Runoff: length from adverse crown removed to full superelevated (DOT B to D), D = SC 4. Fully superelevate remainder of curve and then reverse the process at the CS.

36

Imagehttp://techalive.mtu.edu/modules/module0003/Supere levation.htm

37

Same as point E of GB

With Spirals

38Source: Iowa DOT Standard Road Plans RP-2

With Spirals

Tangent runout (A to B)

39

With Spirals

Removal of crown

40

With Spirals

Transition of superelevation

Full superelevation

41

42

Transition ExampleGiven: PI @ station 245+74.24 D = 4 (R = 1,432.4 ft) ( = 55.417 L = 1385.42 ft

43

With no spiral T = 752.30 ft PC = PI T = 238 +21.94

44

For: Design Speed = 50 mph superelevation = 0.04 normal crown = 0.02 Runoff length was found to be 144 Tangent runout length = 0.02/ 0.04 * 144 = 72 ft.

45

Where to start transition for superelevation? Using 2/3 of Lr on tangent, 1/3 on curve for superelevation runoff: Distance before PC = Lt + 2/3 Lr =72 +2/3 (144) = 168 Start removing crown at: PC station 168 = 238+21.94 - 168.00 = Station = 236+ 53.94

46

Location Example with spiral Speed, e and NC as before and ( = 55.417 PI @ Station 245+74.24 R = 1,432.4 Lr was 144, so set Ls = 150

47

Location Example with spiralSee Iowa DOT design manual for more equations: http://www.dot.state.ia.us/design/00_toc.ht m#Chapter_2 Spiral angle s = Ls * D /200 = 3 degrees P = 0.65 (calculated) Ts = (R + p ) tan (delta /2) + k = 827.63 ft

48

Location Example with spiral TS station = PI Ts = 245+74.24 8 + 27.63 = 237+46.61 Runoff length = length of spiral Tangent runout length = Lt = (eNC / ed ) x Lr = 2% / 4% * 150 = 75 Therefore: Transition from Normal crown begins at (237+46.61) (0+75.00) = 236+71.61

49

Location Example with spiralWith spirals, the central angle for the circular curve is reduced by 2 * s Lc = ((delta 2 * s) / D) * 100 Lc = (55.417-2*3)/4)*100 = 1235.42 ft Total length of curves = Lc +2 * Ls = 1535.42 Verify that this is exactly 1 spiral length longer than when spirals are not used (extra credit for who can tell me why, provide a one-page memo by Monday)50

Location Example with spiralAlso note that the tangent length with a spiral should be longer than the non-spiraled curve by approximately of the spiral length used. (good check but why???)

51

Notes Iowa DOT

Source: Iowa DOT Standard Road Plans Note: Draw a sketch and think about what the last para is saying52


Recommended